Stars in Your Window POJ - 2482

Posted 2020-12-02 hehe54321

错误记录：

题目说输入在int范围内，但是运算过程中可能超int；后来开了很多longlong就过了

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<vector>
  5 #include<map>
  6 using namespace std;
  7 #define fi first
  8 #define se second
  9 #define mp make_pair
 10 #define pb push_back
 11 typedef long long ll;
 12 typedef unsigned long long ull;
 13 typedef pair<int,int> pi;
 14 struct Q
 15 {
 16     ll x1,x2,y;int c,type;
 17 }q[20100];
 18 int nq;
 19 bool operator<(const Q &a,const Q &b)
 20 {
 21     return a.y<b.y||(a.y==b.y&&a.type<b.type);
 22 }
 23 int n,w,h;
 24 ll ans;
 25 namespace S
 26 {
 27 #define lc (num<<1)
 28 #define rc (num<<1|1)
 29 #define mid (l+((r-l)>>1))
 30 ll d[200100],addv[200100];
 31 void build(int l,int r,int num)
 32 {
 33     if(l==r)    {d[num]=addv[num]=0;return;}
 34     build(l,mid,lc);build(mid+1,r,rc);
 35     d[num]=addv[num]=0;
 36 }
 37 int L,R;ll x;
 38 void pd(int l,int r,int num)
 39 {
 40     d[lc]+=addv[num];d[rc]+=addv[num];
 41     addv[lc]+=addv[num];addv[rc]+=addv[num];
 42     addv[num]=0;
 43 }
 44 void _addx(int l,int r,int num)
 45 {
 46     if(L<=l&&r<=R)
 47     {
 48         d[num]+=x;addv[num]+=x;
 49         return;
 50     }
 51     pd(l,r,num);
 52     if(L<=mid)    _addx(l,mid,lc);
 53     if(mid<R)    _addx(mid+1,r,rc);
 54     d[num]=max(d[lc],d[rc]);
 55 }
 56 ll _que(int l,int r,int num)
 57 {
 58     if(L<=l&&r<=R)    return d[num];
 59     pd(l,r,num);
 60     ll ans=0;
 61     if(L<=mid)    ans=max(ans,_que(l,mid,lc));
 62     if(mid<R)    ans=max(ans,_que(mid+1,r,rc));
 63     return ans;
 64 }
 65 #undef lc
 66 #undef rc
 67 #undef mid
 68 }
 69 map<ll,ll> ma;
 70 ll t0[200100];
 71 int main()
 72 {
 73     int i;ll a,b,c;
 74     while(scanf("%d%d%d",&n,&w,&h)==3)
 75     {
 76         nq=0;
 77         for(i=1;i<=n;i++)
 78         {
 79             scanf("%lld%lld%lld",&a,&b,&c);
 80             q[++nq].y=a-w+1;q[nq].x1=b-h+1;q[nq].x2=b;q[nq].c=c;q[nq].type=1;
 81             q[++nq].y=a+1;q[nq].x1=b-h+1;q[nq].x2=b;q[nq].c=c;q[nq].type=0;
 82         }
 83         ans=0;
 84         t0[0]=0;
 85         for(i=1;i<=nq;i++)    t0[++t0[0]]=q[i].x1,t0[++t0[0]]=q[i].x2;
 86         sort(t0+1,t0+t0[0]+1);t0[0]=unique(t0+1,t0+t0[0]+1)-t0-1;
 87         ma.clear();
 88         for(i=1;i<=t0[0];i++)    ma[t0[i]]=i;
 89         for(i=1;i<=nq;i++)    q[i].x1=ma[q[i].x1],q[i].x2=ma[q[i].x2];
 90         sort(q+1,q+nq+1);
 91         S::build(1,t0[0],1);
 92         for(i=1;i<=nq;i++)
 93         {
 94             if(q[i].type==0)
 95             {
 96                 S::L=q[i].x1;S::R=q[i].x2;S::x=-q[i].c;
 97                 S::_addx(1,t0[0],1);
 98             }
 99             else
100             {
101                 S::L=q[i].x1;S::R=q[i].x2;S::x=q[i].c;
102                 S::_addx(1,t0[0],1);
103             }
104             if(i==nq||q[i].y!=q[i+1].y)
105             {
106                 S::L=1;S::R=t0[0];
107                 ans=max(ans,S::_que(1,t0[0],1));
108             }
109         }
110         printf("%lld
",ans);
111     }
112     return 0;
113 }